Group of invertible elements of a Banach algebra ($G(A)$)

The invertible elements of a unital Banach algebra form an open multiplicative group. For $a\in G(A)$, every $b$ satisfying $\Vert a^{-1}\Vert \Vert b-a\Vert <1$ is invertible, because $b=a(1+a^{-1}(b-a))$ and the second factor has a Neumann series inverse.

Finite products of Banach algebra exponentials form a subgroup: inverses are $e^{-a}$ and conjugation sends $e^a$ to $e^{ga{g}^{-1}}$. Each product is joined to the identity by $t\mapsto \prod e^{t{a}_i}$. The logarithm of an element near the identity puts a norm neighborhood of $1$ in this subgroup, so its cosets are open and its complement is open. It is therefore exactly the identity component of a topological group, and is an open, closed, normal subgroup. A product of exponentials need not be a single exponential in a noncommutative algebra.

Suppose $a_n\to a$ in a unital Banach algebra, with each $a_n$ invertible. If $ba=1$, then $b{a}_n\to 1$, so $b{a}_n$ is invertible eventually by the Neumann series. Then $b=(b{a}_n)a_n^{-1}$ is invertible, and $a=b^{-1}$. The analogous argument with $a_{n}b\to 1$ treats a right inverse of an algebra element. Thus a noninvertible such limit admits neither type of one-sided inverse.

A right inverse of an element $a$ is an element $b$ with $ab=1$. A left inverse of an algebra element and a right inverse of an algebra element, when both exist, coincide. For the unilateral shift operator $S$, $S^{\ast }S=1$ but $S{S}^{\ast }\ne 1$, exhibiting the distinction in the Banach algebra of bounded operators.

A left inverse of an element $a$ of a unital associative algebra is an element $b$ with $ba=1$. A right inverse instead satisfies $ab=1$. If both exist, they agree: $b=b(ac)=(ba)c=c$. A one-sided inverse alone need not imply invertibility in an infinite-dimensional Banach algebra.

In a unital Banach algebra, if a subsequence of inverse norms remained bounded, $a=a_n[1+a_n^{-1}(a-a_n)]$ would be invertible for large subsequence indices by the Neumann series. This contradicts noninvertibility of $a$. If $A$ is a closed unital subalgebra of a larger Banach algebra $B$ and the $a_n$ lie in $G(A)$, then such a limit cannot be invertible in $B$ either: continuity of inversion in a Banach algebra would make their inverses converge in $B$ to $a^{-1}$, which closedness puts in $A$.

Inversion is continuous on the group of invertible elements of a Banach algebra. The Neumann series gives a local bound on inverse norms, while $b^{-1}-a^{-1}=b^{-1}(a-b)a^{-1}$ makes the difference tend to zero as $b\to a$.